If an object is moving at #12 m/s# over a surface with a kinetic friction coefficient of #u_k=5 /g#, how far will the object continue to move?

1 Answer
Apr 2, 2017

Answer:

The distance is #=14.4m#

Explanation:

Let the mass of the object be #=mkg#

#v=12ms^-1#

The coefficient of kinetic friction is

#mu_k=F_r/N#

#N=mg#

So,

#F_r=mu_k*N#

#=5/g*mg=5m N#

The work done by #F_r# is equal to the kinetic energy of the object.

#F_r*d=1/2mv^2#

#d=(1/2mv^2)/(5m)=v^2/10=12^2/10=144/10=14.4m#